ELCE 452
Laboratory Report - 1
Solar Panels
Medet Tolepbergenov
Nurtileu Sagyndyq
Date: 04.10.2022
In this laboratory, we worked with the solar panel that had the following characteristics in the
backward sheet:
Table 1. Characteristics of used Solar Panel
Part Number
9046143
Nominal Peak Power (𝑊𝑃 )
100 W
Nominal Voltage (𝑉𝑀𝑃 )
19.55 V
Nominal Current (𝐼𝑀𝑃 )
5.12 A
Open Circuit Voltage (𝑉𝑂𝐶 )
23.15 V
Short Circuit Current (𝐼𝑂𝐶 )
5.45 A
Operating Temperature
-40° to +85°
Maximum System Voltage
1000 V DC
Dimensions: 1055×670×35 mm
Weight: 8 kg
a) Report open circuit voltage for your panel?
From the Table 1, we can see that our panel have an open voltage of: 𝑉𝑂𝐶 = 23.15 V
b) Report short circuit current for your panel?
From the above characteristics Table of our solar panel, it can be seen that the short circuit
voltage is: 𝐼𝑆𝐶 = 5.45 A
c) Report the nominal panel power rate?
According to characteristics of our panel from Table 1, the Nominal Peak Power is:
𝑊𝑃 = 150 W
d) Report the panel output voltage at vertical position, inclined for 0, 45, 90
degree, and also horizontal position? Approximate angle is OK.
According to our measurements, the values of current and voltage at different positions were as
follows:
Table 2. Measured value of voltage and current at different positions
Angle
Voltage (V)
Current (A)
0 degree
12.90
0.007
45 degrees
12.97
0.007
90 degrees
11.81
0.005
e) Can you obtain the maximum power rate from this panel? If so, explain it
how. If not, explain it why?
From our measurements in laboratory condition, we are not able to obtain the maximum power
rate of the panel. We need to have strong perpendicular sunlight rays for our panel which we
could not obtain in the laboratory as it was done in the room. Moreover, to get perpendicular
rays, we need to adjust the angle of the panel as perfectly as possible, which was not the case in
our laboratory. In addition, we need to have a high light intensity, which could be obtained
outside under the sun, however, as we did our measurements in the room, the light intensity was
lower.
f) Suppose you have 10m × 10m roof area, and you want to implement this type
of panels over the roof. How many panel you need to reach the maximum
output power over this roof (in horizontal arrangement)? How much would be
the maximum output power?
From Table 1, we can see that our panel has measurements of 1.055mm × 0.67 mm. Thus, in 10
m × 10 m area we will have:
10/1.055 = 9.47 = 9 panels vertically
10/0.67 = 14.92 = 14 panels horizontally
Thus, in total we will have: 9×14 = 126 panels on 10 m × 10 m roof area.
The illustration for such roof can be seen as follows:
Figure 1. Arrangement of solar panels on 10 m × 10 m roof area
We know from Table 1 that 1 solar panel gives us 100 W of maximum power. Therefore, 126
solar panels will give the maximum power of: 𝑊𝑀𝐴𝑋 = 100 W × 126 = 12.6 kW
Thus, we get 12.6 kW power from 10 m × 10 m roof.
g) With the power obtained in part (f), how many residential apartments can be
supported? Suppose each residential apartment consume 4.1 kW, daily.
So, if every apartment consumes 4.1 kW daily, then by 126 panels we can support:
𝑁 = 12.6 kW / 4.1 kW = 3.07 = 3 apartments.
Thus, by using the power of 126 solar panels, we can support 3 apartments.
h) If the sky becomes cloudy and 45% of the Sun power is lost, how much would
be the output power in part (f)?
When the sky is cloudy, we will only have 55% of our power. Thus, the output power will
become: 𝑃𝑂𝑈𝑇 = 12.6 kW × 0.55 = 6.93 kW.
So, 126 solar panels will give us 6.93 kW power.
i) In part (f), suppose that the roof can tolerate 7 kg/m2 mechanical load.
Suggest the maximum power which we can obtain from this type of panel
over this roof.
From Table 1, we know that our solar panel weighs 8 kg. Thus, we can calculate how much of
our solar panel should be inside of 1 𝑚2 : 7 kg / 8 kg = 0.875.
So, 0.875 of our solar panel should be placed in 1 𝑚2 area. Therefore, we can arrange the solar
panels as follows:
Figure 2. Arrangement of solar panels in roof that can tolerate 7 kg per 𝑚2
The arrangement was done as efficiently as possible. You can see that in the last row, the panels
were placed horizontally in order to create an efficient power. So, after all approximate
arrangements, we had 85 panels that were able to fit on such a roof. That is why, the maximum
power from this roof will be: 100 W × 85 = 8.5 kW
Thus, in total, we can get approximately 8.5 kW of power from such 10 m × 10 m roof.
j) In part (f), if the efficiency of the inverter is 78%, how much would be the
total output power? Is this power enough to support your indicated
apartment in part (g)? If not, give your suggestion for consumers.
In part (f), 12.6 kW of total power was calculated. The 78% will give us:
12.5 kW × 0.78 = 9.828 kW.
To support 3 apartments in part (g), we need to have: 4.1 kW × 3 = 12.3 kW of power which is
larger than 9.828 kW. Thus, we do not have enough power to support 3 apartments with 78%
inverter. In order to have enough power, it is suggested for customers to decrease their power
consumption or the 3rd apartment will need to search for other options of receiving a power.
Moreover, we can also install solar panels that can gain larger power or create a new area with
solar panels in order to support the 3rd apartment.